Answer 3.1
(a) Integration of equation (3.3) yields
(b) Setting T = when x = 0 gives = C. Then, since T(L) = , we have
or
Finally substituting
(3.4)
in the expression for T(x), we obtain
(3.5)
It should be noticed that when TL > To, q is negative; this is consistent with our intuition that heat flows from higher to lower temperatures.
Note too that the temperature distribution is a linear function of x.
Finally, to underscore the role of the gradient in this analysis, it is readily verified that the expressions for and given by equations (3.4) and (3.5) satisfy (1.1), Fourier's law of heat conduction.
(c) To use equation (3.4) to evaluate q, we must employ a consistent set of units. (See the remarks following equation (2.1).) In units of Kelvin,
= 60 + 273 and = 250 + 273, so = 250 - 60 = 190 K. L = 10cm = 0.1m. Therefore
Thus, heat flows through a square meter at the rate of 7,600 watts (or Joules per second). (Recall that a Joule is a unit of energy.)
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